I've been asking questions on reflectors before and I hope you are not getting annoyed. Apologies if that's the case.

My question is the following: Are there reflectors to the forgetful functor $U: \mathbf{CMon} \to \mathbf{Mon}$ from commutative monoids to the general monoids?

I know they exist in rings and groups but I have trouble working it out for monoids. Any answer is very much appreciated, but one not referring to the adjoint functor theorem is preferred.

  • $\begingroup$ Thanks Martin Sleziak for edit, I'll be better in the future $\endgroup$ – user25470 Apr 9 '12 at 20:11

Well it's pretty much the same as for groups, and in fact, every algebraic structure which includes a binary composition law.

When $M$ is a monoid, then $M^{\mathrm{ab}}$ is defined to be $M/\sim$, where $\sim$ is the smallest congruence relation on $M$ which satisfies $ab \sim ba$ for all $a,b \in M$. Then $M \mapsto M^{\mathrm{ab}}$ is left adjoint to the inclusion functor $U$. The proof is trivial.

If you want to have an explicit description of $\sim$ (which is important for computations, but not for the proof that the above is true): $x \sim y$ iff there is a composition $x=x_1 \cdots x_n$ and a permutation $\sigma$ of $1,\dotsc,n$ such that $y = x_{\sigma(1)} \cdots x_{\sigma(n)}$. [This easy description is not available for the category of groups] In other words: Computation in $M^{\mathrm{ab}}$ is as in $M$, but you don't care for the order in which they are done.

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  • $\begingroup$ Does it have a right adjoint? $\endgroup$ – user25470 Apr 9 '12 at 18:44
  • $\begingroup$ No, $U$ doesn't preserve colimits (namely, coproducts are not preserved). $\endgroup$ – Martin Brandenburg Apr 9 '12 at 19:03
  • $\begingroup$ Thanks for the help, deeply appreciated $\endgroup$ – user25470 Apr 9 '12 at 20:10

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